Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 1
Page 7 Problem 1 Answer
Given: 18+9+1+12
To Find: Sum of the given equation
In order to solve the equation, we will divide the equation in two parts and then solve it 18+9+1+12
Lets divide the equation in two parts in order to solve the equation.
=(18+9)+(1+12)
Now, adding the 2 equations separately, we get: =(27)+(13)
To get the final answer, remove the brackets and add the final two digits.
=27+13
=40
The solution of the given equation is 40.
Page 7 Problem 2 Answer
7×15×12
Lets divide the equation in two parts in order to solve the equation.
=(7×15)×2
Now, multiplying the bracket equation, we get:
=(105)×2
To get the final answer, remove the brackets and multiply the final two digits.
=105×2
=210
Given:7×15×2
To Find: Multiplication of the given equation
In order to solve the equation, we will divide the equation in two parts and then solve it.
The solution of the given equation is 210.
Page 7 Problem 3 Answer
Given: 3+41/2+11+51/2
To Find: Sum of the given equation
In order to solve the equation, we have to convert the mixed fraction to improper fraction, converting the improper fraction to like fraction, adding all the like fraction to final answer.
3+41/2+11+51/2
Convert mixed fraction to improper fraction.
[3/1+(4×2)+1/2]+[11/1+(5×2)+1/2]
=[3/1+9/2]+[11/1+11/2]
Convert the L.C.M. of the denominators and change the fraction into like fractions.
(3/1+9/2)+(11/1+11/2)
L.C.M of 1 and 2 is 2.
=(3×2/1×2+9×1/2×1)+(11×2/1×2+11×1/2×1)
=(6/2+9/2)+(22/2+11/2)
Adding the like fraction to get the final answer,
=(6+9/2)+(22+11/2)
=15/2+33/2
=48/2
=24
The Solution to the given equation is 24.
Page 7 Problem 4 Answer
Given:−5×7×20
To Find: Multiply the given equation
In order to solve the equation, we will divide the given equation in two parts and multiply them separately. Since one digit is in negative, then the final answer will also be in negative
−5×7×20
Let’s divide the equation in two parts.
=−5×(7×20)
Now, by multiplying the bracket equation, we get,
−5×(7×20) =−5×(140)
Since, one digit is in negative, we will apply the rule of Negative number×
Positive number=Negative number.
Therefore, the final solution will also be in negative. We will remove the bracket and simply multiply the equation.
=−5×140
−5×140 =−700
The Solution of the given equation is −700.
Page 7 Problem 5 Answer
Given:−12+3+12+19
To Find: Sum of the given equation
In order to solve the equation,We will divide the equation in two partsWe will keep the negative number on the left side and positive on the right sideAdd the positive numbers first and then subtract with the negative number.
The final solution sign should be the same sign as the number with the greater absolute value.
let’s divide the equation in two parts and keep the negative number on the left and positive number on the right.
=(−12)+(3+12+19)
Now, add all the positive numbers:
(−12)+(3+12+19) =(−12)+(34)
Remove the bracket, and then solve the equation. Since12<34, then the final answer will be in positive number. We just need to subtract the final equation.
=−12+34
−12+34 =22
The Solution of the given equation is 22
Page 7 Problem 6 Answer
Given:−1×5×9×2
To Find: Multiply the given equation
In order to solve the equation, we will divide the given equation in two parts and multiply them separately.
Since one digit is in negative, then the final answer will also be in negative.
−1×5×9×2
Let’s divide the equation in two parts. Keep the negative number on the left and positive number on the right.
=(−1)×(5×9×2)
Now, we will multiply the positive number first.
(−1)×(5×9×2) =(−1)×(45×2)
=(−1)×(90)
Since, one digit is in negative, we will apply the rule of Negative number×
Positive number= Negative number.
Therefore, the final solution will also be in negative. We will remove the bracket and simply multiply the equation.
=−1×90
=−90
The Solution of the given equation is −90
Page 7 Problem 7 Answer
Given: 14(12)
To Find: Solution of the equation by distributive property.
In order to solve the equation, we have to remove the bracket and multiply the given digits.
14(12)
In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.
We will multiply the number immediately outside parentheses with those inside:
=14×12
Now we will multiply the equation, and the solution will be:
=168
The Solution of the given equation is 168
Page 7 Problem 8 Answer
Given: 5(47)
To Find: Solution of the equation by distributive property.
In order to solve the equation, we have to remove the bracket and multiply the given digits.
5(47)
In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.
We will multiply the number immediately outside parentheses with those inside:
=5×47
Now we will multiply the equation, and the solution will be:
=235
The solution of the given equation is 235.
Page 7 Problem 9 Answer
Given: 4(106)
To Find: Solution of the equation by distributive property.
In order to solve the equation, we have to remove the bracket and multiply the given digits.
4(106)
In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.
We will multiply the number immediately outside parentheses with those inside:
=4×106
Now we will multiply the equation, and the solution will be:=424
The Solution of the given equation is 424
Page 7 Problem 10 Answer
Given 16x+27x
To Find: Sum of the given equation.
In order to solve the equation, we can combine them into a single term by adding their coefficients.
16x+27x
Let’s combine the equation into a single term by adding their coefficients and taking the x outside the bracket.
=(16+27)x
Add the equation which are in the brackets.=(43)x
Now, remove the bracket to get the final answer.=43x
The Solution of the given equation is 43x.
Page 7 Problem 11 Answer
Given: 6t2−2t2
To Find: Difference of the given equation.
In order to solve the equation, we can combine them into a single term by subtracting their coefficients.
6t2−2t2
Let’s combine the equation into a single term by adding their coefficients and taking the t2 outside the bracket.
=(6−2)t2
Subtract the equation which are in the brackets.=(4)t2
Now, remove the bracket to get the final answer.=4t2
The Solution of the given equation is 4t2.
Page 7 Problem 12 Answer
Given: −5w3+18w3
To Find: Solution of the given equation.
In order to solve the equation, we can combine them into a single term by adding their coefficients.
We have to find the difference of the absolute values by giving the sum the same sign as the number with the greater absolute value.
−5w3+18w3
Let’s combine the equation into a single term by adding their coefficients and taking w3 outside the bracket.
=(−5+18)w3
Since 5<18 then the answer will be in positive number, we just have to subtract the given digits.
=(13)w3
Now, remove the bracket to get the final answer.
=13w3
The Solution of the given equation is 13w3
Page 7 Problem 13 Answer
Given:−2.6d−3.4d
To find: Solution of the equation
In order to solve the equation, we can combine them into a single term by adding their coefficients.
We will also follow the rule of −Negative numbers−Negative numbers =Negative Solution, but we have to add it
−2.6d−3.4d
Let’s combine the equation into a single term by adding their coefficients and taking the outside the bracket.
=(−2.6−3.4)d
When you add two negative integers together, the sum is a more negative number.
=(-6.0)d
Now, remove the bracket to get the final answer.
=−6.0d
The Solution of the given equation is −6.0d.
Page 7 Problem 14 Answer
Given: −12d+3+14d+18
To Find: Solution of the given equation
In order to solve the equation, we can combine them into a single term by adding their coefficients.
We have to find the difference of the absolute values by giving the sum the same sign as the number with the greater absolute value.
−12d+3+14d+18
First, Combine the like terms and divide it with the bracket.
=(−12d+14d)+(3+18)
Add both the equation separately.
Since 12<14 then the answer will be in positive number, we just have to subtract the first equation.
=(2d)+21
Now, remove the bracket to get the final answer.
=2d+21
The solution of the given equation is2d+21.
Page 7 Problem 15 Answer
Given 42x+36x+42x+36x
To Find: Perimeter of the given Parallelogram
In order to solve the equation, we have to add all the sides of the parallelogram.
The given diagram is parallelogram. To know the perimeter we have to add the sides of the parallelogram.
42x+36x+42x+36x
Let’s divide the equation in two parts in order to solve the equation.
=(42x+36x)+(42x+36x)
Now, adding the 2 equations separately, we get:
=(78x)+(78x)
Let’s combine the equation into a single term by adding their coefficients and taking the x outside the bracket.
=(78+78)x
=(156)x
Now, remove the bracket to get the final answer.
=156x
The perimeter of the given parallelogram is 156x.
Page 7 Problem 16 Answer
Given : 3x+4x+8+3(x−2)
To Find: Perimeter of the given diagram.
In order to solve the equation, we have to add all the sides of the triangle.
The given diagram is triangle. To know the perimeter we have to add the sides of the triangle.
3x+4x+8+3(x−2)
Multiply out the brackets and remember that everything inside the brackets will get multiplied by 3
=3x+4x+8+3x−6
Let’s divide the equation in two parts in order to solve the equation.
=(3x+4x+3x)+(8−6)
Now, adding the 2 equations separately, we get:
=10x+2
The perimeter of the given triangle is 10x+2